In exercising, a weight lifter loses 0.200 kg of water through evaporation, the heat required to evaporate the water coming from the weight lifter's body. The work done in lifting weights is 1.50 105 J. (a) Assuming that the latent heat of vaporization of perspiration is 2.42 106 J/kg, find the change in the internal energy of the weight lifter. J (b) Determine the minimum number of nutritional calories of food (1 nutritional calorie

(a) The change in internal energy of the weight lifter is -6.3 × 10⁵ J

(b) The amount of calories of food to be consumed to replace the lost energy is 151.46 calories

Explanation:

Here we have

Mass of water, m = 0.200 kg

Latent heat of vaporization, L = 2.42 × 10⁶ J/kg

Work done in lifting the weight = 1.50 × 10⁵ J

(a) Therefore the heat required to vaporize the 0.200 kg of water is

Q = m×L =  0.200 × 2.42 × 10⁶ J/kg = -‭484,000 J ( -ve as energy is spent)

Therefore, the change in internal energy, ΔU of the weight lifter is;

ΔU  = Q - W = -‭484,000 J - 1.50 × 10⁵ J = ‭-634,000 J = -6.3 × 10⁵ J

(b) The number of calories of food is given by;

1 calorie = 4186 J

Therefore, 6.3 × 10⁵ J contains

151.46 calories should be consumed to make up for the energy used.

i'm thinking more like energy

at positions a, g kinetic energy is maximum.

at position d, potential energy is maximum.